Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be locally Lipschitz and $f(0) = 0$ (so that $x = 0$ is an equilibrium). Consider ODE \begin{equation} \frac{dx(t)}{dt} = f(x(t)), \quad x(0) = x_0 \in \mathbb{R}^n, \quad t \geq 0 \end{equation} and its unique solution $t \mapsto \varphi(t; x_0) \in \mathbb{R}^n$ w.r.t. initial condition $x_0$.
I am concerning about the global existence of the solution on the boundary $\partial B$ of the basin of attraction \begin{equation} B := \big \{ x_0 \in \mathbb{R}^n : \varphi(t; x_0) \text{ exists } \forall t \geq 0 \text{ and } \lim_{t \to \infty} \varphi(t; x_0) = 0 \big \} \end{equation} Since it is well-known that $B$ is open, $\partial B \cap B = \varnothing$. Therefore, it is possible that solution $t \mapsto \varphi(t ; x_0)$ starting at $x_0 \in \partial B$ explodes in finite time and does not exist thereafter (when $\partial B$ is unbounded).
For unbounded $\partial B$, can anyone prove that every solution $t \mapsto \varphi(t ; x_0)$ starting at $x_0 \in \partial B$ does not explode in finite time, or show a counter-example? Many thanks in advance.